How do we identify the structures of the molecules presented in 2443? How do you know the structures of the products I show are correct?

Transcription

1 How do we identify the structures of the molecules presented in 2443? How do you know the structures of the products I show are correct? Can you confirm or refute this for yourself? How to Prove a Structure? 1 Yes! We use spectroscopy. In effect, we hit the molecule with certain types of energy and observe what happens. Mass spectrometry: we obtain the molecular formula Infrared spectroscopy (IR): we determine the functional group Nuclear magnetic resonance spectroscopy (NMR): we establish molecular connectivity

2 energy, in kcal/mol Energy & Light & the Spectrum x10 9 kcal gamma rays 95 kcal 0.95 kcal ultraviolet infra-red 2.86x10 4 kcal x-rays 75 kcal 36 kcal visible 9.5x10-4 kcal 1x10-6 kcal microwaves radio waves cm 10-7 cm 3x10-5 cm 3x10-3 cm 3x10 1 cm 3x10 4 cm x10-5 cm wavelength, in cm

3 Energy & Light & the Spectrum 3 Wavelength and frequency are inversely proportional, and are related by the speed of light c (c = 3x10 10 cm sec -1 ). Frequency is given the symbol ν (in Hz) and wavelength is given the symbol λ (in cm). Energy (E) is proportional to frequency (ν) and related by Planck's constant (h) so that E = h ν. Convert wavelength to nanometers (nm or ν; 1x10-7 cm) Red light appears at 800 nm and violet light at 400 nm. 800 nm light is lower in energy when compared to the 400 nm light. Higher number in nm correlates with lower energy 2.86x10 9 kcal gamma rays energy, in kcal/mol 95 kcal 0.95 kcal ultraviolet infra-red 2.86x10 4 kcal x-rays 75 kcal 36 kcal visible 9.5x10-4 kcal 1x10-6 kcal microwaves radio waves cm 10-7 cm 3x10-5 cm 3x10-3 cm 3x10 1 cm 3x10 4 cm x10-5 cm wavelength, in cm

4 Light and Energy wavelength! longer wavelength, lower energy 4 When the cycle is measured in seconds, one cycle/second is defined a 1 Hertz (Hz). Note that if frequency is one Hz for A and one Hz for B, the wave moves further in one second for A than for B. If we define frequency by distance using centimeters (cm), then one cycle/cm = 1 cm -1. If both A and B show a frequency of 1 cm -1, then it takes more time for A than for B. If radiation is of higher frequency, it will have more waves per second, so the wavelength is shorter. We equate this with higher energy. A B frequency one cycle in one second = 1 Hz wavelength frequency one cycle in one second! lower frequency, fewer cycles per second, lower energy shorter wavelength, higher energy higher frequency, more cycles per second, higher energy LW E: long wavelength, low frequency HIGH E: short wavelength, high frequency

5 Electron Impact Mass Spectrometry Hitting A Molecule With 70 Electron Volts: Electron Impact Mass Spectrometry (mass spec) 5 Bombard a molecule with a high energy electron beam. When this impacts an organic molecule, an electron is ejected to form a radical-cation, which can be detected. This high-energy cation fragments into smaller molecular fragments and these can be separated in individual ions, and then detected. Knowledge of such fragmentations gives structural information. Detection of the molecular radical-cation that has lost only an electron allows us to determine the mass of that fragment and, thereby, of the original molecule. This technique is called mass spectrometry. A radical cation = X P.S. NT Mass spectroscopy

6 Early Mass Spectrometry 6 Early work with mass spectrometers by J.J. Thomsen showed differentiation of isotopes by a stream of ionized neon through a magnetic and electric field, as recorded on photographic plates. Arthur Dempster Thomsen concluded that neon is composed of atoms with two different masses, 20 Ne and 22 Ne. Techniques such as this allowed identification of many naturally occurring isotopes. The first modern mass spectrometer was developed by physicist Arthur Jeffrey Dempster (Canada; USA; ), who discovered the uranium isotope 235 U. The mass spectrometer developed by Dempster allowed identification of compounds by the mass of elements in those compounds, and also the isotope content of the elements. Dempster s work established the basic theory and design of modern mass spectrometers.

7 Mass Spectrometer mass-spectrometer.html Schematic of Electron Impact Mass Spectrometer

8 What results when a molecule is placed in the mass spectrometer? Ejection of an electron to give a high energy radical cation The molecular ion (parent ion) H 3 C CH 3 H 3 C CH 3 mass = 58 mass = 58-1e - Molecular ion or Parent ion H 3 C CH 3 70 ev Mass of acetone 1 e 70 ev Ion Fragmentation: Daughter Ions 8 H 3 C Molecular ion fragments into daughter ions H 3 C CH 3 + H 3 C CH CH High energy + mass = 58 mass = 58-1e - mass = 43 mass = 15

9 Mass Spectrum 9 Separation of all ions, and detection leads to the mass spectrum Abundance is the amount of each ion detected, relative to the ion detected in greatest abundance (base ion = B) Abundance m 3 m 2 m 1 m/z m=0 m=100 m/z = mass/charge mass of each ion divided by the charge of that ion. If the charge is +1, then m/z= mass of the ion

10 Mass spectrum for the ionization of acetone MS of Acetone 10 + Relative Abundance Base ion CH 3 H 3 C + M 15 (loss of methyl) + H 3 C CH 3 Molecular ion m/z It is difficult to identify the structure of ions a priori, but we can Take the parent ion (M or P) and look for the mass of lost fragments. Loss of 15 mass units, for example, corresponds to loss of a methyl group from the parent.

11 Common Fragmentations: M? 11 Fragmentation of the parent can give structural clues P 18 H loss of water M-18 = alcohol P 15 loss of methyl M-15 = parent had a methyl group P 29!-cleavage m/z 29 α-cleavage and M-29 = an ethyl group next to a functional group or heteroatom

12 Actual Mass Spectra 12

13 Determining Molecular Weight of Parent: M, M+1, M+2 The mass of the parent ion should be the highest mass M = molecular ion (also known as P = parent ion) It is not! 13 Ratio of M+1/M and M+2/M M There is a peak with one mass unit higher than M the M+1 Relative Abundance % M % M How? There is a peak with two mass units higher than M the M+2 m/z

14 Common Isotopes 14 Isotope Relative Abundance (%) 12 C C M H H M N N M M M S S M S M Cl Cl M Br Br 98.0 M+2 fluorine ( 19 F), phosphorus ( 31 P), and iodine ( 127 I) do not have M+1 or M+2 isotopes that can be used here.

15 #C, #N and #: Formulas from M+1 & M+2 15 always use the isotope of highest natural abundance to calculate the molecular weight. Since we know the isotopic ratio, measurement of the M+1 peak relative to M should allow us to calculate the number of carbons. This is the basic premise of two formulas that can be used to determine the empirical formula for a molecule. Since the parent ion also provides the molecular weight, if the charge z = +1, we can determine the molecular formula for a sample. Based on the isotope table, the two formulas are: M + 1 = (# C) (1.11) + (# N) (0.38) 13 C 15 N M + 2 = (# C) (1.11) 2 + (# ) (0.20) C 18

16 Even M = Even MW (0 N), dd M = dd MW (1 N): An Assumption 16 Therefore: If M = even mass, assume 0 nitrogen atoms and M+1 = 1.11(#C) If M = odd mass, assume 1 nitrogen atom, and M+1 = 1.11(#C) (1 N) Note: For M+2 to work with oxygen, with only a 0.2 % isotope, Assume no Cl, S or Br. For the time being - but not forever, a molecule will contain S, Cl, or Br, but only one of each and, if the molecule contains S, Cl, or Br, it will not contain. Usually - if there is Br or Cl +, I give you the formula

17 m/z m/z m/z P (162) (100) P+1 (163) (12.21) P+2 (164) (0.95) % % % Example 1 17 C 11 H 14 even mass, so 0 nitrogen, so P+1 = = 1.11(#C) #C = 12.21/1.11 = 11 C 11 P+2 = 0.95 = [(1.11)(11)] 2 / # 0.95 = # # = / 0.2 = 1 1 Cannot see H in mass spec, therefore obtain H by difference from the parent ion (162) 11 C x 12 = x 16 = = 14, so 14 H????? Remember the alkane formula C n H 2n+2 = maximum number of H for a molecule For C11, C 11 H 24, so there cannot be more than 24 H, but there can be less.